$f(x, y) = \sin(x) + \cos(2y)$ What are all the critical points of $f$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $\left( \pi + 2\pi k, \pi j \right)$, where $k, j = \ldots -1, 0, 1 \ldots$ (Choice B) B $\left( \dfrac{\pi}{2} + \pi k, \dfrac\pi 2 j \right)$, where $k, j = \ldots -1, 0, 1 \ldots$ (Choice C) C $\left( \dfrac\pi 2 k, 2\pi j \right)$, where $k, j = \ldots -1, 0, 1 \ldots$ (Choice D) D There are no critical points.
A critical point of a scalar field $f$ is where $\nabla f = \bold{0}$. [What's that bolded 0?] Let's find the gradient of $f$ ! $\nabla f = \begin{bmatrix} \cos(x) \\ \\ -2\sin(2y) \end{bmatrix}$ We want each component of the gradient to equal zero, so we want to solve the system of equations below. $\begin{cases} \cos(x) = 0 \\ \\ -2\sin(2y) = 0 \end{cases}$ Let's start by solving $\cos(x) = 0$. There are many values for $x$ that satisfy this equation. For example, $x = \dfrac{\pi}{2}$ makes $\cos(x) = 0$. Note that $\cos(x)$ is generally periodic every $2\pi$, but it crosses the $x$ -axis every $\pi$. So we can add any integer multiple of $\pi$ to get a new solution. All solutions will have the form $x = \dfrac{\pi}{2} + \pi k$, for some integer $k$. Now let's solve $-2\sin(2y) = 0$. We can simplify this to $\sin(2y) = 0$. There are many values for $y$ that satisfy this equation. For example, $y = 0$ makes $\sin(2y) = 0$. Note that $\sin(2y)$ is generally periodic every $\pi$, but it crosses the $x$ -axis every $\dfrac\pi 2$. So we can add any integer multiple of $\dfrac\pi 2$ to get a new solution. All solutions will have the form $y = \dfrac\pi 2 j$, for some integer $j$. Therefore, $f$ has critical points at $\left( \dfrac{\pi}{2} + \pi k, \dfrac\pi 2 j \right)$, where $k, j = \ldots -1, 0, 1 \ldots$.